What is the tangent of y=(36)/(x^2+2),(1,12) ?

The tangent of y=(36)/(x^2+2),(1,12) is y=-8x+20

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tangent of y=(36)/(x^2+2),(1,12)

tangent of

y = \frac{36}{x ^{2} + 2}, (1, 12)

y = - 8 x + 20

Solution steps

Find the slope of

y = \frac{36}{x ^{2} + 2} : \frac{d y}{d x} = - \frac{72 x}{( x ^{2} + 2 ) ^{2}}

Compute the slope of

y = \frac{36}{x ^{2} + 2}

(1, 12) : m = - 8

Find the line with slope m=

- 8

and passing through

(1, 12) : y = - 8 x + 20

y = - 8 x + 20

\frac{\partial}{\partial x} (\frac{a x - 5}{- y ^{2} + 1})

x \to 0 + lim (arctan (x))

\int \frac{x ^{4}}{4 + x ^{2}} d x

\int_{- 2}^{3} x - 4 ∣ x ∣ d x

\frac{1}{x ^{2}}

What is the tangent of y=(36)/(x^2+2),(1,12) ?
The tangent of y=(36)/(x^2+2),(1,12) is y=-8x+20